Let $R,T>0$ and $u\in C([0,T)\times[0,R])C^{1,\:2}((0,T)\times[0,R])$ satisfy $$u_t=u_{xx}+u^p\;\;\;\text{in }(0,T)\times(-R,R)\tag1$$ for some $p>1$ and $$u=0\;\;\;\text{on }(0,T)\times\{-R,R\}.\tag2$$ Now let $\alpha\in(1,p)$.
Question 1: How can we show that $$\int_0^xu^{-\alpha}(t,r)u_r(t,r)\:{\rm d}r=\frac{u^{1-\alpha}(t,x)-u^{1-\alpha}(t,0)}{1-\alpha}\tag3$$ for all $x\in[0,R]$?
I've tried to obtain the right-hand side using partial integration, but wasn't able to simplify the resulting question in a way which yields the right-hand side. Is there something I'm missing which yields the desired equality in a different way?
Assume now that $$u(0,\;\cdot\;)=u_0\;\;\;\text{on }[-R,R]$$ for some $u_0:[-R,R]\to[0,\infty)$ such that $u_0\not\equiv0$, $u_0$ is nonincreasing on $[0,R]$ and satisfies $$u_0(x)=u_0(-x)\;\;\;\text{for all }x\in[-R,R]\tag4.$$
Question 2: How do we see that $u^{-\alpha}(t,x)$ is actually well-defined for all $t\in[0,T)$ and $x\in[0,R)$?
We obviously need to exclude $x=R$, since $u(t,R)=0$ for all $t\in(0,T)$ by $2$. But how do we now that $u(t,x)\ne0$ for all $t\in[0,T)$ and $x\in[0,R)$?
As Arctic Char mentioned, even though it is partial derivatives you can think of it as similar to saying: $$\int yy' dx=\int ydy$$ Since you know that: $$\partial_r u^{1-\alpha}=(1-\alpha)u^{-\alpha}\partial_ru=(1-\alpha)u^{-\alpha}u_r$$ which means you can rewrite your integral as: $$\int_0^x\partial_r\left[\frac{u^{1-\alpha}}{1-\alpha}\right]dr$$ which will give the desired result
One comment that you can make is that by rearranging the equation we get: $$u=(u_t-u_{xx})^{1/p}$$ so for $u$ to be real (unless $p$ is an odd integer) we require $u_t>u_{xx}$.
One thing to note is that from the conditions you have stated we can say that $u_0'\le0\,\,\forall x\in[0,R]$ and obviously symmetrical about $x=0$ for this to be true we can say one of two things, $u_0'(x)=0$ or $u_0'(0)=0$. Think about what the behaviour of $u_0$ tells you about $u$